Problem: $f(x)=-2x^4-2x^3+60x^2-22$. On which intervals is the graph of $f$ concave down? Choose 1 answer: Choose 1 answer: (Choice A) A $x>2$ only (Choice B) B $x<\dfrac{5}{2}$ and $x>5$ (Choice C) C $-\dfrac{5}{2}<x<2$ only (Choice D) D $x<-\dfrac52$ and $x>2$
Explanation: We can analyze the intervals where $f$ is concave up/down by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding increasing/decreasing intervals, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=-12(2x+5)(x-2)$. $f''(x)=0$ for $x=-\dfrac52,2$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our points of interest are $x=-\dfrac52$ and $x=2$. Our points of interest divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x<\ -\frac{5}{2}$ $-\frac{5}{2}<x<2$ $x>2$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<-\dfrac{5}{2}$ $x=-3$ $f''(-3)=-60<0$ $f$ is concave down $\cap$ $-\dfrac{5}{2}<x<2$ $x=0$ $f''(0)=120>0$ $f$ is concave up $\cup$ $x>2$ $x=3$ $f''(3)=-132<0$ $f$ is concave down $\cap$ In conclusion, the graph of $f$ is concave down over the intervals $x<-\dfrac52$ and $x>2$.